Question

The number of ordered pairs $ (x, y) $ for which $ A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 2 & x \\ y & 1 & 2 \end{pmatrix} $ is a singular and symmetric matrix is:

• A. \( 1 \)
• B. \( 0 \)
• C. \( 2 \)
• D. \( 3 \)

Hint:

To satisfy both symmetry and singularity: - First apply symmetry conditions to reduce the number of variables. - Then check if the determinant is zero with those values.

Answer 1

The Correct Option is B

Step 1: Use the condition that matrix \(A\) is symmetric.
A matrix is symmetric if \( A = A^T \), i.e., \( a_{ij} = a_{ji} \). So for matrix \(A\), this gives: \[ A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 2 & x \\ y & 1 & 2 \end{pmatrix} \Rightarrow x = 1 \quad \text{and} \quad y = 1 \] Step 2: Substitute \(x = 1\), \(y = 1\) into matrix \(A\): \[ A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix} \] Step 3: Use the condition that \(A\) is singular.
A matrix is singular if \( \det(A) = 0 \). Compute the determinant: \[ \det(A) = 1 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} \] \[ = 1(4 - 1) - 2(4 - 1) + 1(2 - 1) = 3 - 6 + 1 = -2 \neq 0 \] So, the matrix is not singular for \(x = 1\), \(y = 1\).
Conclusion: There are no values of \(x, y\) that make the matrix both symmetric and singular.